Problem: Solve the equation. $\dfrac{dy}{dx}=\dfrac{x}{\sin(y)}-\dfrac{9}{\sin(y)}$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=\arccos\left(-\dfrac{x^2}{2}+9x+C\right)$ (Choice B) B $y=\arccos\left(-\dfrac{x^2}{2}+9x\right)+C$ (Choice C) C $y=\dfrac{2}{\cos\left(-{x^2}+18x+C\right)}$ (Choice D) D $y=\dfrac{2C}{\cos\left(-{x^2}+18x\right)}$
Explanation: We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{x}{\sin(y)}-\dfrac{9}{\sin(y)} \\\\ \dfrac{dy}{dx}&=\dfrac{1}{\sin(y)}(x-9) \\\\ \sin(y)\,dy&=(x-9)\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} \sin(y)\,dy&=(x-9)\,dx \\\\ \int \sin(y)\,dy&=\int (x-9)\,dx \\\\ -\cos(y)&=\dfrac{x^2}{2}-9x+C_1 \\\\ \cos(y)&=-\dfrac{x^2}{2}+9x+C \\\\ \arccos(\cos(y))&=\arccos\left(-\dfrac{x^2}{2}+9x+C\right) \\\\ y&=\arccos\left(-\dfrac{x^2}{2}+9x+C\right) \end{aligned}$ [Where did we get C?] Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=\arccos\left(-\dfrac{x^2}{2}+9x+C\right)$